SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
Quantum Rotation in 3 Dimensions (Rigid Rotor)
A model of a rotating diatomic molecule is created as follows:
 Assume that two masses, m1 and m2, are connected to one another
by a rigid rod of length R. System is called a rigid rotor.
 This system is allowed to freely rotate in 3D space about its center
of mass. Since the distance between the masses is unvarying, we
call this the rigid rotor model.
 Potential energy of this system is 0.
𝑚 𝑚
 𝜇 = 𝑚 1+𝑚2 is defined to be the reduced mass.
1
2
 𝐼 = 𝜇𝑅 2 is defined to be the moment of inertia.
 This system is best represented in spherical polar coordinates,
with the radius r=R is fixed.
Derivation of the Wave Function
The wave function is stated (using the reduced mass ) in spherical coordinates with the Laplacian
highlighted:
ℏ2
− 2𝜇 ∇2 Ψ(r, θ, ϕ) + 𝑉(r, θ, ϕ)Ψ(r, θ, ϕ) = 𝐸Ψ(r, θ, ϕ).
(1)
The following stipulations are applied to the potential energy:
0 𝑟=𝑅
𝑉(r, θ, ϕ) = {
.
∞ 𝑟≠𝑅
(2)
Therefore:
−
ℏ2 2
∇ Ψ(r, θ, ϕ)
2𝜇
= 𝐸Ψ.
(3)
Expanding the Laplacian in terms of spherical coordinates and dropping the independent variables
from Ψ yields:
ℏ2
1 ∂
− 2𝜇 [𝑟2 ∂𝑟 (𝑟 2
∂Ψ
1
∂
∂Ψ
1
∂Ψ
) + 𝑟2 sin(𝜃) ∂𝜃 (sin(𝜃) ∂𝜃 ) + 𝑟2 sin2 (𝜃) ∂𝜙2 ]
∂𝑟
= 𝐸Ψ.
Three things to note here:


Recall that the derivation of the Laplacian in polar coordinates was quite
tedious. The tedium is tenfold for the derivation of the Laplacian in spherical
coordinates, so it is merely stated above without derivation. 
In physics and chemistry, the definitions of the angles 𝜃 and 𝜙 have are
switched with respect to the definitions you have learning in mathematics. 𝜃
1
(4)
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)

measures the angle off of the positive z-axis, and 𝜙 is the counter clockwise angle
measurement off of the positive x-axis. This distance from the origin is represented by r
vice 𝜌.
Since the radius is constant (i.e. particles will not exist in a regime where the potential
energy is infinite) we can let 𝑟 = 𝑅 and
∂Ψ
∂𝑟
= 0 which allows us to simplify the wave
equation: 
ℏ2
1
∂
∂Ψ
1
∂Ψ
− 2𝜇𝑅2 [sin(𝜃) ∂𝜃 (sin(𝜃) ∂𝜃 ) + sin2 (𝜃) ∂𝜙2 ] = 𝐸Ψ.
(5)
Using the definition of the moment of inertia 𝐼 = 𝜇𝑅 2, this becomes:
ℏ2
1
∂
∂Ψ
1
∂Ψ
− 2𝐼 [sin(𝜃) ∂𝜃 (sin(𝜃) ∂𝜃 ) + sin2(𝜃) ∂𝜙2 ] = 𝐸Ψ.
(6)
ℏ2
Dividing both sides by − 2𝐼 yields:
1
∂
∂Ψ
(sin(𝜃) ∂𝜃 )
sin(𝜃) ∂𝜃
1
∂Ψ
+ sin2 (𝜃) ∂𝜙2 = −
2𝐼𝐸
Ψ.
ℏ2
(7)
Note:
∂
∂Ψ
(sin(𝜃) ∂𝜃 )
∂𝜃
∂2 Ψ
∂Ψ
= cos(𝜃) ∂𝜃 + sin(𝜃) ∂𝜃2 .
(8)
Therefore:
∂2 Ψ
cos(𝜃) ∂Ψ
1
∂Ψ
+
+ 2
∂𝜃2
sin(𝜃) ∂𝜃
sin (𝜃) ∂𝜙2
=−
∂2 Ψ
cos(𝜃) ∂Ψ
1
∂Ψ
+ sin(𝜃) ∂𝜃 + sin2 (𝜃) ∂𝜙2
∂𝜃2
+
2𝐼𝐸
Ψ,
ℏ2
(9)
and finally,
2𝐼𝐸
Ψ
ℏ2
= 0.
(10)
We now separate the variables, by letting let Ψ(θ, ϕ) = Θ(θ)Φ(ϕ)(or more simply written: Ψ =
ΘΦ. ) This implies that:
∂Ψ
∂𝜃
= Θ′Φ,
∂2 Ψ
∂𝜃2
= Θ′′Φ, and
∂2 Ψ
∂𝜙2
= ΘΦ′′.
(11)
Thus:
Θ′′Φ +
cos(𝜃)
1
2𝐼𝐸
Θ′Φ + 2 ΘΦ′′ + 2 ΘΦ
sin(𝜃)
sin (𝜃)
ℏ
Dividing equation by ΘΦ yields
2
= 0.
(12)
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
Θ′′ cos(𝜃) Θ′
1 Φ′′ 2𝐼𝐸
+
+ 2
+ 2 = 0.
Θ
sin(𝜃) Θ sin (𝜃) Φ
ℏ
Multiply by sin2 (𝜃):
sin2 (𝜃)
Θ′′
Θ′ Φ′′ 2𝐼𝐸sin2 (𝜃)
+ sin(𝜃)cos(𝜃) +
+
= 0.
Θ
Θ
Φ
ℏ2
Separating variables yields:
sin2(𝜃)
Θ′′
Θ′ 2𝐼𝐸 sin2 (𝜃)
Φ′′
+ sin(𝜃) cos(𝜃) +
=
−
= 𝓂2 ,
Θ
Θ
ℏ2
Φ
where M represents the eigenvalues. We now have two ordinary differential equations:
Φ′′ + 𝓂2 Φ = 0
Θ′′ +
cos(𝜃)
2𝐼𝐸
𝓂2
Θ′ + ( 2 − 2 ) Θ = 0
sin(𝜃)
ℏ
sin (𝜃)
The solution to the first equation is familiar:
Φ𝓂 (ϕ) = c1 ei𝓂ϕ .
Recall the derivation from the particle in a ring problem. After applying boundary conditions (i.e.
Φ(ϕ + 2π) = Φ(ϕ)) we noted that 𝓂 must be an integer. After normalization conditions are
applied the solution is:
Φ𝓂 (ϕ) =
1
ei𝓂ϕ .
√2𝜋
To solve the second equation we let 𝑥 = cos(𝜃):
dΘ dΘ dx
dΘ
dΘ
=
= − sin(𝜃)
= −√1 − 𝑥 2 ,
d𝜃 d𝑥 d𝜃
d𝑥
d𝑥
which implies:
d
d
= −√1 − 𝑥 2 .
d𝜃
d𝑥
We use this to find:
d2 Θ
𝑑 dΘ
∂
dΘ
d2 Θ
dΘ
2)
2
2
√
√
(1
=
(
)
=
−
1
−
𝑥
(−
1
−
𝑥
)
=
−
𝑥
−
𝑥
d𝜃 2 𝑑𝜃 d𝜃
∂𝑥
d𝑥
d𝑥 2
d𝑥
3
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
Substituting this into:
d2 Θ cos(𝜃) dΘ
2𝐼𝐸
𝓂2
+
+
−
(
) Θ = 0,
d𝜃 2 sin(𝜃) d𝜃
ℏ2 sin2 (𝜃)
Yields:
(1 − 𝑥
2)
d2 Θ
dΘ
𝑥
dΘ
2𝐼𝐸
𝓂2
2
√
−𝑥
+
(− 1 − 𝑥
)+( 2 −
) Θ = 0,
d𝑥 2
d𝑥 √1 − 𝑥 2
d𝑥
ℏ
1 − 𝑥2
which simplifies to:
(1 − 𝑥 2 )
Let
2𝐼𝐸
=ℓ(ℓ −
ℏ2
d2 Θ
dΘ
2𝐼𝐸
𝓂2
−
2𝑥
+
−
(
)Θ = 0
d𝑥 2
d𝑥
ℏ2
1 − 𝑥2
1). This gives us:
(1 − 𝑥 2 )
d2 Θ
dΘ
𝓂
− 2𝑥
+ (ℓ(ℓ − 1) −
) Θ = 0,
2
d𝑥
d𝑥
1 − 𝑥2
which is known as the associate Legendre differential equation. The equation has non-singulars
solutions for 𝑥 ∈ (−1,1) only if ℓ is an integer and 0 ≤ |𝓂| ≤ ℓ. The integers ℓ and 𝓂 are
respectively referred to as the degree and order of the associate Legendre function. Note that 𝑥 =
cos(𝜃) so the singularity condition is met.
4
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
The derivation of a final solution is not discussed here and left open for exercise. The solutions to
|𝓂|
the general Legendre equation are called the associated Legendre functions 𝑃ℓ
examples are given below:
Associated Legendre Functions for 𝓵 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒
𝓵 |𝓶| 𝑷𝓶
𝓵 (𝒙)
0
1
0
0
1
0
2
1
2
0
3
1
2
3
ℓ
|𝓂|
𝑃00 = 1
𝑃10 = 𝑥
𝑃11 = −√1 − 𝑥 2
1
𝑃20 = (3𝑥 2 − 1)
2
1
𝑃2 = −3𝑥√1 − 𝑥 2
𝑃22 = 3(1 − 𝑥 2 )
1
𝑃30 = (5𝑥 3 − 3𝑥)
2
3
𝑃31 = − (5𝑥 2 − 1)√1 − 𝑥 2
2
𝑃32 = 15𝑥(1 − 𝑥 2 )
𝑃33 = −15(1 − 𝑥 2 )3/2
1
𝑑ℓ+|𝓂|
𝓂
2 |𝑚|/2
𝑃ℓ = ℓ (1 − 𝑥 )
(1 − 𝑥 2 )ℓ
2 ℓ!
𝑑𝑥 ℓ+|𝓂|
5
(𝑥). Some
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
Recall that we had let 𝑥 = cos(𝜃). Back substituting, the above table is converted to the context of
the rigid rotor problem:
Associated Legendre Functions for 𝓵 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒
𝓵 |𝓶| 𝑷|𝓶| (𝒄𝒐𝒔(𝜽))
𝓵
0
1
0
0
1
0
2
1
2
0
3
1
2
3
𝑃00 = 1
𝑃10 = cos(𝜃)
𝑃11 = −sin(𝜃)
1
𝑃20 = (3cos2 (𝜃) − 1)
2
𝑃21 = 3 cos(𝜃) sin(𝜃)
𝑃22 = −3sin2 (𝜃)
1
𝑃30 = (5cos3 (𝜃) − 3cos(𝜃))
2
3
𝑃31 = − (5cos2 (𝜃) − 1)sin(𝜃)
2
𝑃32 = 15 cos(𝜃) sin2 (𝜃)
𝑃33 = −15sin3 (𝜃)
We now have a solution for Ψ(θ, ϕ):
Ψℓ,𝓂 (θ, ϕ) =
1
|𝓂|
A
P (cos(θ))ei𝓂ϕ.
√2𝜋 ℓ,|𝓂| ℓ
To find Aℓ,|𝓂| , a double integral is used to normalize Ψℓ,𝓂 (θ, ϕ):
0
2𝜋
1
|𝓂|
∫𝜋 ∫0 |√2𝜋 Aℓ,|𝓂| Pℓ
2
(cos(θ))ei𝓂ϕ | 𝑑ϕ 𝑑𝜃.
Since ϕ and 𝜃 are independent in this expression, the double integral can be expressed as the
product of two integrals:
0
|𝓂|
(A2ℓ,|𝓂| ∫𝜋 |Pℓ
2
2𝜋 i𝓂ϕ 2
1
|e
| 𝑑ϕ)=1.
∫
0
2𝜋
√
(cos(θ))| sin(θ)𝑑𝜃) (
We showed earlier that the second integral equals 1. Therefore:
0
|𝓂|
A2ℓ,|𝓂| ∫ |Pℓ
2
(cos(θ))| sin(θ)𝑑𝜃 = 1
𝜋
We note here that the associate Legendre functions form an orthogonal set and that for 0 ≤ 𝓂 ≤ ℓ
the follow is true:
6
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
2(ℓ + |𝓂|)!
∫ 𝑃𝓀𝓂 (𝑥)𝑃ℓ𝓂 (𝑥)𝑑𝑥 = {(2ℓ + 1)(ℓ − |𝑚|)!
1
𝓀=ℓ
−1
0
𝓀≠ℓ
Since 𝑥 = 𝑐𝑜𝑠(𝜃) (thus 𝑑𝑥 = − sin(𝜃)𝑑𝜃)we rewrite this expression as:
2(ℓ + |𝓂|)!
(2ℓ
+ 1)(ℓ − |𝑚|)!
∫ 𝑃𝓀𝓂 (cos(𝜃))𝑃ℓ𝓂 (cos(𝜃))sin(𝜃) 𝑑𝜃 = {
𝜋
𝓀=ℓ
0
0
𝓀≠ℓ
We can no conclude that:
2(ℓ+|𝓂|)!
A2ℓ,|𝓂| ((2ℓ+1)(ℓ−|𝑚|)!) = 1,
Which implies that:
Aℓ,|𝓂| = √
(2ℓ+1)(ℓ−|𝑚|)!
2(ℓ+|𝓂|)!
.
We now have a solution for the wave function:
Ψℓ,𝓂 (θ, ϕ) = √
(2ℓ+1)(ℓ−|𝑚|)!
4𝜋(ℓ+|𝓂|)!
|𝓂|
Pℓ
(cos(θ))ei𝓂ϕ.
Note that earlier we let:
2𝐼𝐸
=ℓ(ℓ − 1)
ℏ2
And stated that ℓ must be an integer in order to find non-singular solutions to the Legendre
function. Thus energy is quantized:
𝐸=
ℓ(ℓ−1)ℏ2
2𝐼
=
ℓ(ℓ−1)ℏ2
2𝜇𝑅2
=
ℓ(ℓ−1)(𝑚1 +𝑚2 )ℏ2
.
2𝑚1 𝑚2 𝑅2
Each integer ℓ has 2ℓ + 1 values of 𝓂 associated with it that do not affect the energy levels. Thus
each energy level has associated with it 2ℓ + 1 wave forms.
7
SM286 – Spring 2010
Supplementary Notes 12
Quantum Rotation (3D Rigid Rotor)
Example: Find the wave form associated with ℓ = 2 and 𝓂 = 1:
(5)(1)! 1
5
Ψ2,1 (θ, ϕ) = √
P (cos(θ))eiϕ = √
3 cos(𝜃) sin(𝜃) eiϕ =
4𝜋(3)! 2
24𝜋
45
15
√15
√
cos(𝜃) sin(𝜃) eiϕ = √ cos(𝜃) sin(𝜃) eiϕ =
cos(𝜃) sin(𝜃) eiϕ
24𝜋
8𝜋
2√2𝜋
8